Question

The velocity time relation of a particle starting from the rest is given by V = KL where K = 2 m/s² . Calculate The distance travelled by a particle in 3 sec .​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Given :-

a = 2 m/s²

Which is k so we put 2 m/s² in the place of a

${\boxed{\sf\:{Using\;3rd\;equation\;of\;motion}}}$

v² = u² + 2as

v² = (0)² + 2as

v² = 2as

${\boxed{\sf\:{We\;can\;also\;write\;it\;as}}}$

$\tt{\rightarrow s=\dfrac{v^2}{2a}}$

As V = KL

Put KL in place of V

L = Time (t)

$\tt{\rightarrow s=\dfrac{(KL)^2}{2K}}$

$\tt{\rightarrow s=\dfrac{K^{2}L^{2}}{2K}}$

$\tt{\rightarrow s=\dfrac{KL^{2}}{2}}$

$\tt{\rightarrow s=\dfrac{1}{2}\times KL^{2}}$

$\tt{\rightarrow s=\dfrac{1}{2}\times 2\times (3)^{2}}$

s = 9 metre

The distance travelled by a particle in 3 sec is 9 metres.

Answer 2

$\mathfrak{\huge{\pink{ANSWER}}}$

Given ,

$\boxed{Velocity\:Time\:Relation}$

v = kt

Given k = 2 m/sec2 ,

t = 3 sec

We acceleration , 

a  = dv / dt

= d (kt )/dt

= k = 2 m/sec2

As we know ,

s=ut + 1/2×a× t2

s=1/2×2× 32

s=9m

The distance travelled by a particle in 3 sec is 9 m.